On the Weak Index Problem for Game Automata

Abstract

Game automata are known to recognise languages arbitrarily high in both the non-deterministic and alternating Rabin–Mostowski index hierarchies. Recently it was shown that for this class both hierarchies are decidable. Here we complete the picture by showing that the weak index hierarchy is decidable as well. We also provide a procedure computing for a game automaton an equivalent weak alternating automaton with the minimal index and a quadratic number of states. As a by-product we obtain that, as for deterministic automata, the weak index and the Borel rank coincide.

The authors have been supported by Foundation for Polish Science grant Homing Plus no. 2012-5/1.

Appendices

A Upper Bounds

Lemma 3

If $\mathrm {wclass}(\mathcal {A}, q)\le \mathbf{{RM}}^{w}(i,j)$, one can construct effectively a weak alternating automaton of index $(i,j)$ with $\mathcal {O} (|Q^\mathcal {A} |^2)$ states, recognising $L(\mathcal {A}, q)$.

The algorithm never returns $\mathbf {\Pi }^{w}_{0} = \mathbf{{RM}}^{w}(0,0)$ nor $\mathbf {\Sigma }^{w}_{0} =\mathbf{{RM}}^{w}(1,1)$, so the lowest $(i, j)$ to consider are (0, 1) and (1, 2). Assume $(i, j)=(0,1)$. Examining the algorithm we see that this happens only if no rejecting loop is reachable from state q. Since automaton $\mathcal {A}$ is priority-reduced, it means that $\mathcal {A}$ uses only priority 0. Hence, it is already a (0, 1) weak automaton (not (0, 0), because of possible $\bot$ transitions). For $(i, j)=(1,2)$ the argument is entirely analogous.

For higher indices we consider three cases, leading to three different constructions of weak alternating automata recognising $L(\mathcal {A},q)$.

$\mathcal {B}$ has no $\forall$ -branching transitions, $(i,j)=(1,j)$, $j\ge 3$. In an initial part of the weak automaton recognising $L(\mathcal {A},q)$ the players declare whether during the play on a given tree they would leave component $\mathcal {B}$ or not. Since $\mathcal {B}$ has no $\forall$-branching transitions, as long as the play stays in $\mathcal {B}$, each choice of $\forall$ amounts to leaving $\mathcal {B}$ or staying in $\mathcal {B}$. Hence, each strategy of $\exists$ admits exactly one path staying in $\mathcal {B}$, finite or infinite. We first let $\exists$ declare $l_\exists \in \{\textit{leave},\textit{stay}\}$: $\textit{leave}$ means that the path is finite, $\textit{stay}$ means that it is infinite.

• If $l_\exists =\textit{leave}$, we move to a copy of $\mathcal {B}$ with all the priorities set to 1. By Eq. (2), for every exit f of $\mathcal {B}$ we have $\mathrm {wclass}(\mathcal {A},f)\le \mathbf{{RM}}^{w}(1,j)$. Therefore, we can compose this copy of $\mathcal {B}$ with all the automata for $L(\mathcal {A}, f)$ to obtain an automaton of index $(1,j)$, and we are done.

• Assume that $l_\exists =\textit{stay}$. Given the special shape of $\exists$’s strategies, this means that $\exists$ claims that the play will only leave $\mathcal {B}$ if at some point $\forall$ chooses an exit f in a transition whose other end is in $\mathcal {B}$. Since the minimal priority in $\mathcal {B}$ is 0, all these exists are $(\forall ,0)$-replicated. We check $\exists$’s claim by substituting all other exits in transitions with rejecting states, i.e. weak alternating automata of index (3, 3) (recall that $j$ is at least 3). Thus, the only exits that are not substituted are the $(\forall ,0)$-replicated ones.

Now, assuming $l_\exists =\textit{stay}$, we ask $\forall$ whether he plans to take one of these exists: he declares $l_\forall \in \{\textit{leave},\textit{stay}\}$, accordingly.

• If $l_\forall =\textit{stay}$, the play moves to the weak alternating automaton of index , corresponding to the co-deterministic automaton $\mathcal {B}$ with the remaining exits removed from transitions (they were only present in transitions of the form $(q_\mathtt {L},\mathtt {L}) \wedge (q_\mathtt {R},\mathtt {R})$, with the other state in $\mathcal {B}$).

• Assume that $l_\forall =\textit{leave}$. In that case we move to a copy of $\mathcal {B}$ with all the priorities set to 2. The only exits left are the $(\forall ,0)$-replicated ones. By Eq. (2), for such exists f, $\mathrm {wclass}(\mathcal {A}, f)\le \mathbf{{RM}}^{w}(0,j-2)$: otherwise $\mathrm {wclass}(\mathcal {A}, f)\ge \mathbf{{RM}}^{w}(1,j-1)$, so $\big (\mathrm {wclass}(\mathcal {A}, p)\big )^\forall \ge \mathbf{{RM}}^{w}(0,j-1)$ and $\mathbf{{RM}}^{w}(0,j-1)$ is not smaller than $\mathbf{{RM}}^{w}(1,j)$. In particular, we can find a weak alternating automaton of index $(2,j)$ recognising $L(\mathcal {A}, f)$. So the whole subautomaton is a weak alternating automaton of index $(2,j)$.

$\mathcal {B}$ has no $\forall$ -branching transitions, $(i,j)=(0,j)$, $j\ge 2$. The simulation starts in a copy of $\mathcal {B}$ with all the priorities set to 0. If the play leaves $\mathcal {B}$ at this stage then we move to the appropriate automaton of index $(0,j)$. At any moment $\forall$ can pledge that:

• the play will no longer visit transitions $\delta (q',a)$ of the form $(f_\mathtt {L},\mathtt {L})\wedge (f_\mathtt {R},\mathtt {R})$, $(f_\mathtt {L},\mathtt {L})\vee (f_\mathtt {R},\mathtt {R})$, $(q_\mathtt {L},\mathtt {L})\vee (f_\mathtt {R},\mathtt {R})$, $(f_\mathtt {L},\mathtt {L})\vee (q_\mathtt {R},\mathtt {R})$, or $(q_\mathtt {L},\mathtt {L})\vee (q_\mathtt {R},\mathtt {R})$, where $\max _{\varOmega }(q_\mathtt {L}\rightarrow q')=\max _{\varOmega }(q_\mathtt {R}\rightarrow q')=0$ and $f_\mathtt {L}$, $f_\mathtt {R}$ are exits of $\mathcal {B}$;

• in the transitions he controls, he will always choose the state in $\mathcal {B}$, and win regardless of $\exists$’s choices.

If the play stays forever in $\mathcal {B}$ but $\forall$ is never able to make such a pledge, he loses by the parity condition — it means that infinitely many times a loop from $q_\mathtt {L}\rightarrow q'$ or $q_\mathtt {R}\rightarrow q'$ is taken with $\max _\varOmega (q_d\rightarrow q')=0$ therefore, the minimal priority occurring infinitely often is 0.

After $\forall$ has made the above pledge, $\exists$ has the following choices:

• She can challenge the first part of $\forall$’s pledge, declaring that at least one such transition is reachable. In that case we move to a copy of $\mathcal {B}$ with all the priorities set to 1 and all the transitions controlled by $\exists$. In this copy, reaching any of the disallowed transitions entails acceptance—the play immediately moves to a (2, 2) final component.

• She can accept the first part of $\forall$’s pledge.

After $\exists$ has accepted the first part of $\forall$ ’s pledge, we can assume that the rest of the game in $\mathcal {B}$ is a single infinite branch. Indeed, by the hypothesis of the theorem, for every $\exists$-branching transition $\delta (q',a)=(q_\mathtt {L},\mathtt {L})\vee (q_\mathtt {R},\mathtt {R})$ in $\mathcal {B}$ it must hold that $\max _{\varOmega }(q_\mathtt {L}\rightarrow q')=\max _{\varOmega }(q_\mathtt {R}\rightarrow q')=0$; otherwise, $\mathcal {B}$ would contain a dual split. Thus, no $\exists$-branching transition can be reached, and since $\mathcal {B}$ contains no $\forall$-branching transitions at all, the game can continue in $\mathcal {B}$ in only one way.

Now $\exists$ must challenge the second part of $\forall$’s pledge. We ask her whether she plans to leave $\mathcal {B}$ or not, and she declares $l_\exists \in \{\textit{leave},\textit{stay}\}$.

• If $l_\exists =\textit{stay}$ then we proceed to the weak automaton of index $\mathrm {wclass}(\mathcal {B},q)$, corresponding to $\mathcal {B}$ treated as a co-deterministic automaton. We are only interested in the behaviour of this automaton over trees in which there is exactly one branch in $\mathcal {B}$, and it is infinite. Over such trees we want to make sure that neither players ever chooses to exit. This is already ensured: when $\mathcal {B}$ is turned into a co-deterministic tree automaton, the exits are simply removed from transitions (if both states are exits, the transition is changed to a transition to a (2, 2) automaton, but such transitions will never be used over trees we are interested in).

• If $l_\exists =\textit{leave}$ then we move to a copy of $\mathcal {B}$ with all the priorities set to 1. The only available exits of $\mathcal {B}$ in this copy are those in transitions of the form $\delta (q',q) = (q_\mathtt {L},\mathtt {L})\vee (f,\mathtt {R})$ (or symmetrical) with $\max _{\varOmega }(q_\mathtt {L}\rightarrow q')>0$ (in other transitions the exits are removed, if both states are exits, they are replaced by a final (2, 2)-component); therefore $\mathrm {wclass}(\mathcal {A}, f)\le \mathbf{{RM}}^{w}(1,j)$ and we can simulate it with a $(1,j)$-automaton.

$\mathcal {B}$ contains $\forall$ -branching transitions If $\mathcal {B}$ contains an $\forall$-branching transition, the algorithm returns $\mathrm {wclass}(\mathcal {A}, q)$ of the form $\mathbf{{RM}}^{w}(0,j)$. Let us construct a weak automaton of index $(0,j)$ that recognises $L(\mathcal {A},q)$. The automaton starts in a copy of $\mathcal {B}$ with all the priorities set to 0. At any moment $\forall$ can declare that no-one will ever take any bad transition in $\mathcal {B}$. If he cannot make such a declaration, it means that $\exists$ can force infinitely many bad transitions to be taken, and she wins. After $\forall$ has made such declaration, we need to recognise the language $L(\mathcal {A} ^{-},q)$ (note that the bad transitions in $\mathcal {A} ^{-}$ are made directly losing for $\forall$). For this we can use a weak automaton of index $\mathrm {wclass}(\mathcal {A} ^{-})\le \mathbf{{RM}}^{w}(0,j)$, already constructed.

Constructed automaton has quadratic number of states. The preprocessing we make to guarantee that the automaton is priority-reduced does not increase the number of states. The resulting automaton consists of:

• a fixed number of copies of $\mathcal {B}$,

• a weak alternating automaton of index ,

• a fixed number of states where players make decisions (e.g. $l_\forall \in \{\textit{leave},\textit{stay}\}$),

• inductively constructed automata recognizing $L(\mathcal {A},f)$ for all exists f of $\mathcal {B}$.

By [Mur08, Theorem 5.5], the automaton in the second item has $\mathcal {O} (|Q^\mathcal {B} |^2)$ states. Hence, we inductively ensure the constructed automaton has $\mathcal {O} (|Q^\mathcal {A} |^2)$ states.

When $\mathcal {A}$ is priority-reduced with all the states productive, the rest of the construction is polynomial in the number of states of $\mathcal {A}$. Therefore, the whole construction can be done in the time of solving the emptiness and completeness problems of $L(\mathcal {A},q)$ for each state q of $\mathcal {A}$ separately.

B Lower Bounds

Lemma 2

If $\mathrm {wclass}(\mathcal {A}, q)\ge \mathbf{{RM}}^{w}(i,j)$ then $S_{(i,j)}\le _{\mathrm {W}}L(\mathcal {A}, q)$.

We prove this claim by induction on the structure of the DAG of SCCs of $\mathcal {A}$ reachable from q, following the cases of the algorithm just like for the upper bound. One of the cases is covered by the procedure for deterministic automata, which we use as a black box. But in order to prove Lemma 1 we need to know that it preserves our invariant. And indeed, just like here, it is a step in the correctness proof: if the procedure returns at least $\mathbf{{RM}}^{w}(i,j)$, then $S_{(i,j)}$ continuously reduces to the recognised language [Mur08]. The remaining cases essentially correspond to the items in Lemma 4 (below).

Lemma 4

Assume that q is a state of $\mathcal {A}$, $\mathcal {B}$ is the SCC of $\mathcal {A}$ containing q, and p is a state of $\mathcal {A}$ reachable from q (from the same or different SCC).

• 1. $L(\mathcal {A},p)\le _{\mathrm {W}}L(\mathcal {A},q)$.

• 2. $L(\mathcal {A} ^-,q)\le _{\mathrm {W}}L(\mathcal {A},q)$.

• 3. If an accepting loop is reachable from q, then $S_{(0,1)} \le _{\mathrm {W}}L(\mathcal {A},q)$.

• 4. If a rejecting loop is reachable from q, then $S_{(1,2)} \le _{\mathrm {W}}L(\mathcal {A},q)$.

• 5. If p is $(\forall ,0)$-replicated by $\mathcal {B}$ then $\left( L(\mathcal {A}, p)\right) ^\forall \le _{\mathrm {W}}L(\mathcal {A}, q)$.

• 6. If p is $(\exists ,1)$-replicated by $\mathcal {B}$ then $\left( L(\mathcal {A}, p)\right) ^\exists \le _{\mathrm {W}}L(\mathcal {A}, q)$.

Proof

The proof is based on the following observation. Let $t\in \mathrm {PTr}_{{A}}$ be a partial tree and $\rho =\rho (\mathcal {A}, t, q_I)$ be the run of an automaton $\mathcal {A}$ on t. We say that t resolves $\mathcal {A}$ from $q_I\in Q^\mathcal {A}$ if $\rho (h) \ne *$ for each hole h of t and whenever $t\!\upharpoonright _{vd}$ is the only total tree in $\{t\!\upharpoonright _{v\mathtt {L}}, t\!\upharpoonright _{v\mathtt {R}}\}$, either $\rho (vd) = *$ or ${\mathbf {G_\rho }}(\mathcal {A}, t\!\upharpoonright _{vd}, \rho (vd))$ is losing for the owner of v. Assume that a tree t with a single hole h resolves $\mathcal {A}$ from $q_I$ and take $\rho =\rho (\mathcal {A}, t, q_I)$. The notion of resolving is designed precisely so that $t[h:=s]\in L(\mathcal {A}, q_I)$ iff $s\in L(\mathcal {A}, \rho (h))$ for all $s\in \mathrm {Tr}_{{A}}$.

Let us begin with (1). Since all the states of $\mathcal {A}$ are non-trivial, we can construct a tree t with a hole h such that t resolves $\mathcal {A}$ from q and the state $\rho (\mathcal {A}, t, q)(h)$ is p. In that case $t[h:=s]\in L(\mathcal {A},q)$ if and only if $s\in L(\mathcal {A}, p)$. Therefore, the function $s\mapsto t[h:=s]$ is a continuous reduction witnessing that $L(\mathcal {A},p)\le _{\mathrm {W}}L(\mathcal {A},q)$.

For (2), recall that $\mathcal {A} ^-$ is obtained from $\mathcal {A}$ by turning some choices for $\forall$ to $\top$; that is, some transitions $\delta (q',a)$ of the form $(q_\mathtt {L}, \mathtt {L}) \wedge (q_\mathtt {R}, \mathtt {R})$ are set to $(q_\mathtt {L}, \mathtt {L})$, $(q_\mathtt {R}, \mathtt {R})$, or $\top$. This means that if a node v of tree t has label a and gets state $q'$ in the associated run $\rho (\mathcal {A} ^-, t, q)$, then $t\!\upharpoonright _{v\mathtt {L}}$, $t\!\upharpoonright _{v\mathtt {R}}$, or both of them, respectively, are immediately accepted by $\mathcal {A} ^-$. In the corresponding run of the original automaton $\mathcal {A}$, however, these subtrees will be inspected by the players and we should make sure they are accepted. Since $q_\mathtt {L}$ and $q_\mathtt {R}$ are non-trivial in $\mathcal {A}$, we can do it by replacing these subtrees with $t_{q_\mathtt {L}} \in L(\mathcal {A},q_\mathtt {L})$, or $t_{q_\mathtt {R}} \in L(\mathcal {A},q_\mathtt {R})$, accordingly. This gives a continuous reduction of $L(\mathcal {A} ^-,q)$ to $L(\mathcal {A},q)$.

To prove (3), let us fix a state p on an accepting loop C, reachable from q. By (1) and transitivity of $\le _{\mathrm {W}}$, it is enough to show that $S_{(0,1)}\le _{\mathrm {W}}L(\mathcal {A},p)$. Let t be a tree with hole h such that t resolves $\mathcal {A}$ from p, the state $\rho (\mathcal {A},t,p)$ is p, and the states on the shortest path from the root to h correspond to the accepting loop C. Since all states in $\mathcal {A}$ are non-trivial, we can also find a full tree $t' \notin L(\mathcal {A}, p)$. Let $t_0=t'$ and $t_n = t[h:=t_{n-1}]$ for $n>0$, and let $t_{\infty }$ be the tree defined co-inductively as

$$\begin{aligned} t_{\infty } = t[h:=t_{\infty }]\,. \end{aligned}$$

Then, $t_n\notin L(\mathcal {A},p)$ for all $n\ge 0$, but $t_{\infty } \in L(\mathcal {A},p)$. To get a continuous function reducing $S_{(0,1)}$ to $L(\mathcal {A},p)$, map tree $s\in \mathrm {Tr}_{{\{\bot ,\top \}}}$ to $t_m$, where $m = \min \left\{ i \bigm | s(\mathtt {L}^i\mathtt {R})=\bot \right\}$, or to $t_{\infty }$ if $\left\{ i \bigm | s(\mathtt {L}^i\mathtt {R})=\bot \right\}$ is empty. Item (4) is analogous.

For (5), let us assume that $\delta (q,a)=(q_\mathtt {L},\mathtt {L})\wedge (p,\mathtt {R})$ is the transition witnessing that p is $(\forall ,0)$-replicated by $\mathcal {A}$. Let us also fix the path $q_\mathtt {L}\rightarrow q$ with minimal priority 0. Now, let t be a tree with a hole h that resolves $\mathcal {A}$ from q and the value of the run of $\mathcal {A}$ in h is q. Similarly, let $t'$ be the tree with a hole $h'$ that resolves $\mathcal {A}$ from $q_\mathtt {L}$ and the value of the respective run is q. Let us construct a continuous function that reduces $\left( L(\mathcal {A}, p)\right) ^\forall$ to $L(\mathcal {A}, q)$. Assume that a given tree s has subtrees $s_i$ under the nodes $\mathtt {L}^i \mathtt {R}$. Let us define co-inductively $t_i$ as

$$\begin{aligned} t_i=a(t'[h':=t_{i+1}], s_i), \end{aligned}$$

i.e. the tree with the root labelled by a and two subtrees: $t'[h':=t_{i+1}]$ and $s_i$. Finally, let f(s) be $t[h:=t_0]$. Note that the run $\rho (\mathcal {A}, f(s), q)$ labels the hole h of t by $q'$. Therefore, $f(t)\in L(\mathcal {A},q)$ if and only if $t_0\in L(\mathcal {A}, q')$ and $t_i\in L(\mathcal {A}, q)$ if and only if $t_{i+1}\in L(\mathcal {A},q)$ and $s_i\in L(\mathcal {A},p)$. Since the minimal priority on the path from $t_i$ to $t_{i+1}$ is 0, if no $s_i$ belongs to $L(\mathcal {A}, p)$ then $f(t)\notin L(\mathcal {A}, q)$. Therefore, f is in fact the desired reduction. The proof of (6) is entirely analogous.    $\Box$

Using Lemma 4, and the guarantees for deterministic automata discussed earlier, we prove Lemma 2 as follows.

Proof of Lemma 2. By induction on the recursion depth of the algorithm execution we prove that if $\mathrm {wclass}(\mathcal {A},p)\ge \mathbf{{RM}}^{w}(i,j)$ then $S_{(i,j)}\le _{\mathrm {W}}L(\mathcal {A}, p)$.

Let us start with the lowest level. Assume that $(i,j)=(0,1)$ (for (1, 2) the proof is analogous). Examining the algorithm we see that this is only possible if there is an accepting loop in $\mathcal {A}$, reachable from q. Then, by Lemma 4 Item (3), $S_{(0,1)} \le _{\mathrm {W}}L(\mathcal {A},q)$.

For higher levels we proceed by case analysis. First we cover the possible reasons why Eq. (2) can give at least $\mathbf{{RM}}^{w}(i,j)$. If , the invariant follows immediately from the guarantees for deterministic automata, and the duality between indices and between Skurczyński languages. If $\mathrm {wclass}(\mathcal {A},p)\ge \mathbf{{RM}}^{w}(i,j)$ for some $p\in F$, we use the fact that $L(\mathcal {A},p)\le _{\mathrm {W}}L(\mathcal {A},q)$, and get $S_{(i,j)}\le _{\mathrm {W}}L(\mathcal {A},q)$ by transitivity. Then, assume that $\mathrm {wclass}(\mathcal {A},p)^\exists \ge \mathbf{{RM}}^{w}(i,j)$ for some $p\in F_{\exists ,1}$ (for $p\in F_{\forall ,0}$ the proof is analogous). That means that $\mathrm {wclass}(\mathcal {A},p)\ge \mathbf{{RM}}^{w}(0,j')$ such that $\left( \mathbf{{RM}}^{w}(0,j')\right) ^\exists = \mathbf{{RM}}^{w}(1,j'+2) \ge \mathbf{{RM}}^{w}(i,j)$. By the inductive hypothesis $S_{(0,j')}\le _{\mathrm {W}}L(\mathcal {A}, p)$, so by the monotonicity of $\exists$ and Lemma 4 Item (6), $S_{(1,j'+2)} = \left( S_{(0,j')}\right) ^\exists \le _{\mathrm {W}}\left( L(\mathcal {A},p)\right) ^\exists \le _{\mathrm {W}}L(\mathcal {A},q)$. But since $\mathbf{{RM}}^{w}(1,j'+2) \ge \mathbf{{RM}}^{w}(i,j)$, by the Wadge ordering of Skurczyński’s languages $S_{(i,j)} \le _{\mathrm {W}}S_{(1,j'+2)}$, and $S_{(i,j)} \le _{\mathrm {W}}L(\mathcal {A},q)$ follows by transitivity.

Finally, assume that $\mathrm {wclass}(\mathcal {A},q)$ is computed according to (3); that is, the component $\mathcal {B}$ contains an $\forall$-branching transition $\delta (q',a) = (q_\mathtt {L},\mathtt {L})\wedge (q_\mathtt {R},\mathtt {R})$. As we have already observed, the hypothesis of the theorem implies that in that case $\max _\varOmega (q_\mathtt {L}\rightarrow q') = 0$ and $\max _\varOmega (q_\mathtt {R}\rightarrow q') \le 1$ (or symmetrically). That means that $q_\mathtt {R}$ is $\forall ,0$-replicated by $\mathcal {B}$, so by Lemma 4 Item (5), $\left( L(\mathcal {A}, q_\mathtt {R})\right) ^\forall \le _{\mathrm {W}}L(\mathcal {A}, q)$. But since $\mathcal {B}$ is strongly connected, q is reachable from $q_\mathtt {L}$ and $q_\mathtt {L}$, so by Lemma 4 Item (1) we have $L(\mathcal {A}, q)\le _{\mathrm {W}}L(\mathcal {A}, q_\mathtt {R})$. Since $\forall$ is monotone, we conclude that

$$\begin{aligned} \left( L(\mathcal {A}, q)\right) ^\forall \le _{\mathrm {W}}L(\mathcal {A}, q)\,. \end{aligned}$$

(4)

(It looks paradoxical, but note that $(L^\forall )^\forall \le _{\mathrm {W}}L^\forall$ for all L.) As $\mathrm {wclass}(\mathcal {A}, q) \ge \mathbf{{RM}}^{w}(i,j)$, it must hold that $\mathrm {wclass}(\mathcal {A} ^-, q) \ge \mathbf{{RM}}^{w}(1,j')$ for some $j'$ such that $\left( \mathbf{{RM}}^{w}(1,j')\right) ^\forall = \mathbf{{RM}}^{w}(0,j')\ge \mathbf{{RM}}^{w}(i,j)$. By the induction hypothesis, $S_{(0,j')}\le _{\mathrm {W}}L(\mathcal {A} ^-, q)$. Consequently, by Lemma 4 Item (2) and transitivity, $S_{(0,j')}\le _{\mathrm {W}}L(\mathcal {A}, q)$. Since the operation $\forall$ is monotone, (4) gives $S_{(0,j')} = \left( S_{(1,j')}\right) ^\forall \le _{\mathrm {W}}L(\mathcal {A}, q)$, and we conclude by the $\le _{\mathrm {W}}$ ordering of Skurczyński’s languages.    $\Box$